# What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?

Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules. Then $\mathfrak g\text{-rep}$ comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over $\mathbb C$. Moreover, $\mathfrak g\text{-rep}$ is symmetric monoidal with duals, and the forgetful functor preserves all this structure. By Tannaka-Krein duality (see in particular the excellent paper André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991), from this data we can reconstruct an affine algebraic group $\mathcal G$ such that $\mathfrak g \text{-rep}$ is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of $\mathcal G$.

However, it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group. So it is not true that $\mathcal G$ is, say, necessarily the simply-connected connected Lie group with Lie algebra $\mathfrak g$, or some quotient thereof. So my question is:

Given $\mathfrak g$, what is an elementary description of $\mathcal G$ (that avoids the machinery of Tannaka-Krein)?

For example, perhaps $\mathcal G$ is some Zariski closure of something...?

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Very nice question. – Mariano Suárez-Alvarez Apr 15 '10 at 4:09
It might be something very big since that is so even for some Lie algebras which are Lie algebras of algebraic groups. For example, let $\mathfrak{g}$ be a 1-dimensional Lie algebra. Even though it is the Lie algebra of an algebraic group it seems to me that $\mathcal{G}$ in this case is infinite dimensional. – ulrich Apr 15 '10 at 5:05
Since representations of g can be identified with modules over U(g) perhaps the question should be generalized to Hopf algebras? – Qiaochu Yuan Apr 15 '10 at 5:56
There seems to be an assumption implicit in the question that when you start with an algebraic Lie algebra g, the affine group scheme G you get has Lie algebra g. As @unknown noted, this is far from true (except for semisimple Lie algebras in characteristic zero). So before trying to understand the affine group schemes you get from nonalgebraic Lie algebras, perhaps you should try to understand those you get from algebraic Lie algebras (e.g., the one-dimensional Lie algebra). – JS Milne Apr 15 '10 at 6:15
@QY: Absolutely. But notice that for Tannaka-Krein to reconstruct an actual group puts restrictions on the category of modules: it must be symmetric, and the forgetful functor must preserve the symmetry. From the Hopf algebraic perspective, this is the requirement that H = U(g) is cocommutative, and then TK will reconstruct some dual to H, which will be a commutative algebra and hence a scheme. But anyway, if H is cocommutative, it's pretty close to a universal enveloping algebra already. – Theo Johnson-Freyd Apr 15 '10 at 15:51

Instead of some exotic nonalgebraic Lie algebra, let's start with the one-dimensional Lie algebra $\mathfrak{g}$ over a field $k$. A representation of $\mathfrak{g}$ is just a finite-dimensional vector space + an endomorphism. I don't know what the affine group scheme attached to this Tannakian category is, but thanks to a 1954 paper of Iwahori, I can tell you that its Lie algebra can be identified with the set of pairs $(\mathfrak{g},c)$ where $\mathfrak{g}$ is a homomorphism of abelian groups $k\to k$ and $c$ is an element of $k$. So if $k$ is big, this is huge; in particular, you don't get an algebraic group. (Added: $k$ is algebraically closed.)

By contrast, the affine group scheme attached to the category of representations of a semisimple Lie algebra $\mathfrak{g}$ in characteristic zero is the simply connected algebraic group with Lie algebra $\mathfrak{g}$.

In summary: this game works beautifully for semisimple Lie algebras (in characteristic zero), but otherwise appears to be a big mess. See arXiv:0705.1348 for a few more details.

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Not that I disagree with the conclusion of this answer but the group scheme is $\mathrm{Spec}k[t]\times\mathrm{Spec}k[k]$ with $t\mapsto 1\otimes t+t\otimes1$ and $k[k]$ is the group algebra of the additive group k (and thus $\lambda\mapsto\lambda\otimes\lambda$). (I think that both for this and for the Lie algebra description one has to assume that $k$ is algebraically closed.) – Torsten Ekedahl Apr 15 '10 at 8:57
Thanks --- that looks like the correct way to interpret Iwahori's statements (and, yes, I should have assumed k to be algebraically closed). – JS Milne Apr 15 '10 at 9:11

After some thought my pessimism (as expressed in my concurrence with the answer of Milne) has abated somewhat. If I were bold enough I would conjecture the following (assuming that the characteristic zero base field is algebraically closed): Let $\mathfrak g$ be a finite dimensional Lie algebra over $k$ and let $G$ be the pro-algebraic group whose representation tensor category is equivalent to the tensor category of finite dimensional $\mathfrak g$-modules. Then if $S$ is the (pro-)radical of $G$ and $U$ the (pro-)unipotent radical $U$ and $G/S$ are algebraic groups (unipotent and semi-simple respectively). Furthmore, the pro-torus $T:=S/U$ has as character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ considered as an additive group. Hence the only infinite-dimensional part is $T$ but its character group, \emph{à priori} only an abstract group, is reasonably well controlled. This is analogous to the case of irreducible infinite-dimensional representations of a semi-simple Lie group where the center of the enveloping algebra acts by a character and the set of characters as a set is very large. However it is the set of $k$-points of an algebraic variety which means that it is under control. The analogy goes further as the category of $G$-representations (assuming $U$ is finite dimensional) splits up into a direct product of categories parametrised by cosets of the character group of $T$ with respect to the subgroup generated by the characters occurring in the action of $T$ on the Lie algebra of $U$.

Here are some comments on the conjecture (I do not vouch for the complete veracity of my claims).

We can get a picture of $G/U$ by looking at the irreducible $\mathfrak g$-representations (as they correspond exactly to the irreducible $G/U$-representations). All such representations factor through $\mathfrak g/[\mathfrak{g},\mathfrak{u}]$ which is the product of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and $\mathfrak g/\mathfrak{u}$. Hence, the irreducible representations are parametrised by pairs of a $1$-dimensional representation of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and an irreducible representation of the semi-simple algebra $\mathfrak g/\mathfrak{u}$. This gives the prediction that $G/U$ should be the product of a torus with character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and the simply-connected semi-simple group with Lie algebra $\mathfrak g/\mathfrak{u}$.

As for $U$, the idea is that the category of unipotent representations (i.e., successive extensions of the trivial representation) of $\mathfrak g$ is equivalent to the category of representations of the unipotent group with Lie algebra $\mathfrak g'$, the maximal unipotent quotient of $\mathfrak g$. Something similar ought to be true for successive extension of the same irreducible representation and there shouldn't be too much "intermixing" between different irreducibles.

[Added] I somewhat rudely hijacked the question by taking up things that maybe weren't that pertinent to the question so let me give an answer which I think is more on track.

The problem is that one can not always define the algebraisation of an abstract finite dimensional Lie algebra $\mathfrak g$ even if some algebraisation exists. As an examples consider a $2$-dimensional Lie algebra with basis $x,y$ and $[x,y]=y$. This is the Lie algebra of an infinite number of algebraic groups: Let the $1$-dimensional torus $\mathbb G_m$ act on the additive group $\mathbb G_a$ by $(t,v) \mapsto t^nv$, where $n\not=0$ and let $G_n$ be the semi-direct product of this action. These groups all have $\mathfrak g$ as Lie algebra but the only isomorphisms between them is that $G_n$ is isomorphic to $G_{-n}$.

What does make sense is to speak of an algebraic hull of an embedding of $\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak g$-representation. In that case one may consider the intersection of all algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak g$. In terms of Zariski closures (when the base field is $\mathbb C$) it is the Zariski closure of the exponentials of all elements of $\mathfrak g$ (inside of $\mathfrak{gl}_m$). From the Tannakian point of view this is the group that corresponds to the tensor subcategory of the category of $\mathfrak g$-representations generated by the given representation.

However, if one wants something that is independent of a particular representation one has to pass to an inverse limit of groups coming from different representations. This leads to an infinite dimensional monster even in the case when $\mathfrak g$ is $1$-dimensional.

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Was there some point above at which $\mathfrak{g}$ switched from the finite dimensional Lie algebra we had in the beginning to the infinite dimensional Lie(G)? – S. Carnahan Apr 15 '10 at 15:35
No, I don't think so. The intent is certainly that is the same throughout. I do however switch freely back and forth between $G$-representations and $\mathfrak g$-representations as their categories are equivalent. – Torsten Ekedahl Apr 15 '10 at 18:38